Eve is organizing a race to raise money for a local children's hospital. She makes $20 for each 5 K participant and $45 for each 10 K participant. Her goal is to raise $9,500.
If [tex]$x$[/tex] represents the number of participants in the 5 K race and [tex]$y$[/tex] represents the number of participants in the 10 K race, the linear equation for the given scenario is [ ] [tex]$x+$[/tex] [ ] [tex]$y=$[/tex] [ ].
If Eve raises exactly $9,500, is it possible for there to be 220 participants in the 5 K race? [ ] (yes/no)
Answer (2)
Form the linear equation: 20 x + 45 y = 9500 .
Substitute x = 220 into the equation: 20 ( 220 ) + 45 y = 9500 .
Solve for y : y = 45 5100 = 3 340 ≈ 113.33 .
Since y must be an integer, it is not possible to have 220 participants in the 5K race. n o
Explanation
Forming the Linear Equation Let's break down this problem step by step. First, we need to form a linear equation that represents the total amount Eve raises from the race. We know that each 5K participant contributes $20 and each 10K participant contributes $45, and her goal is to raise $9,500.
The Linear Equation If x is the number of 5K participants and y is the number of 10K participants, the total amount raised can be represented as: 20 x + 45 y = 9500 This is the linear equation for the given scenario.
Substituting x = 220 Now, we need to determine if it's possible for there to be 220 participants in the 5K race. To do this, we substitute x = 220 into the equation: 20 ( 220 ) + 45 y = 9500
Simplifying Simplify the equation: 4400 + 45 y = 9500
Isolating 45y Subtract 4400 from both sides: 45 y = 9500 − 4400
45 y = 5100
Solving for y Now, solve for y by dividing both sides by 45: y = 45 5100 y = 3 340 y ≈ 113.33
Checking for Integer Solution Since the number of participants must be an integer, and y = 3 340 is not an integer (it's approximately 113.33), it is not possible for there to be exactly 220 participants in the 5K race if Eve raises exactly $9,500.
Final Answer Therefore, the linear equation is 20 x + 45 y = 9500 , and it is not possible to have 220 participants in the 5K race.
Examples
This type of problem is useful in budgeting and planning events. For example, if a school is organizing a fundraising event with different ticket prices, they can use a similar equation to determine how many tickets of each type they need to sell to reach their fundraising goal. By understanding these relationships, organizers can make informed decisions to achieve their financial targets.
The linear equation representing the situation is 20 x + 45 y = 9500 . If there are 220 participants in the 5K race, this leads to a non-integer value for the number of 10K participants, meaning it is not possible. Therefore, the answer is no.
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